AVL with duplicate keys

Last Updated : 14 Feb, 2023

Please refer below post before reading about AVL tree handling of duplicates.

How to handle duplicates in Binary Search Tree?
This is to augment AVL tree node to store count together with regular fields like key, left and right pointers.
Insertion of keys 12, 10, 20, 9, 11, 10, 12, 12 in an empty Binary Search Tree would create following.

          12(3)
       /        \
     10(2)      20(1)
    /    \       
 9(1)   11(1)

Count of a key is shown in bracket

Below is implementation of normal AVL Tree with count with every key. This code basically is taken from code for insert and delete in AVL tree. The changes made for handling duplicates are highlighted, rest of the code is same.

The important thing to note is changes are very similar to simple Binary Search Tree changes.

C++

// C++ program of AVL tree that
// handles duplicates
#include <bits/stdc++.h>
using namespace std;
 
// An AVL tree node
struct node {
    int key;
    struct node* left;
    struct node* right;
    int height;
    int count;
};
 
// A utility function to get maximum of two integers
int max(int a, int b);
 
// A utility function to get height of the tree
int height(struct node* N)
{
    if (N == NULL)
        return 0;
    return N->height;
}
 
// A utility function to get maximum of two integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
/* Helper function that allocates a new node with the given key and
    NULL left and right pointers. */
struct node* newNode(int key)
{
    struct node* node = (struct node*)
        malloc(sizeof(struct node));
    node->key = key;
    node->left = NULL;
    node->right = NULL;
    node->height = 1; // new node is initially added at leaf
    node->count = 1;
    return (node);
}
 
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct node* rightRotate(struct node* y)
{
    struct node* x = y->left;
    struct node* T2 = x->right;
 
    // Perform rotation
    x->right = y;
    y->left = T2;
 
    // Update heights
    y->height = max(height(y->left), height(y->right)) + 1;
    x->height = max(height(x->left), height(x->right)) + 1;
 
    // Return new root
    return x;
}
 
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct node* leftRotate(struct node* x)
{
    struct node* y = x->right;
    struct node* T2 = y->left;
 
    // Perform rotation
    y->left = x;
    x->right = T2;
 
    // Update heights
    x->height = max(height(x->left), height(x->right)) + 1;
    y->height = max(height(y->left), height(y->right)) + 1;
 
    // Return new root
    return y;
}
 
// Get Balance factor of node N
int getBalance(struct node* N)
{
    if (N == NULL)
        return 0;
    return height(N->left) - height(N->right);
}
 
struct node* insert(struct node* node, int key)
{
    /* 1.  Perform the normal BST rotation */
    if (node == NULL)
        return (newNode(key));
 
    // If key already exists in BST, increment count and return
    if (key == node->key) {
        (node->count)++;
        return node;
    }
 
    /* Otherwise, recur down the tree */
    if (key < node->key)
        node->left = insert(node->left, key);
    else
        node->right = insert(node->right, key);
 
    /* 2. Update height of this ancestor node */
    node->height = max(height(node->left), height(node->right)) + 1;
 
    /* 3. Get the balance factor of this ancestor node to check whether
       this node became unbalanced */
    int balance = getBalance(node);
 
    // If this node becomes unbalanced, then there are 4 cases
 
    // Left Left Case
    if (balance > 1 && key < node->left->key)
        return rightRotate(node);
 
    // Right Right Case
    if (balance < -1 && key > node->right->key)
        return leftRotate(node);
 
    // Left Right Case
    if (balance > 1 && key > node->left->key) {
        node->left = leftRotate(node->left);
        return rightRotate(node);
    }
 
    // Right Left Case
    if (balance < -1 && key < node->right->key) {
        node->right = rightRotate(node->right);
        return leftRotate(node);
    }
 
    /* return the (unchanged) node pointer */
    return node;
}
 
/* Given a non-empty binary search tree, return the node with minimum
   key value found in that tree. Note that the entire tree does not
   need to be searched. */
struct node* minValueNode(struct node* node)
{
    struct node* current = node;
 
    /* loop down to find the leftmost leaf */
    while (current->left != NULL)
        current = current->left;
 
    return current;
}
 
struct node* deleteNode(struct node* root, int key)
{
    // STEP 1: PERFORM STANDARD BST DELETE
 
    if (root == NULL)
        return root;
 
    // If the key to be deleted is smaller than the root's key,
    // then it lies in left subtree
    if (key < root->key)
        root->left = deleteNode(root->left, key);
 
    // If the key to be deleted is greater than the root's key,
    // then it lies in right subtree
    else if (key > root->key)
        root->right = deleteNode(root->right, key);
 
    // if key is same as root's key, then This is the node
    // to be deleted
    else {
        // If key is present more than once, simply decrement
        // count and return
        if (root->count > 1) {
            (root->count)--;
            return NULL;
        }
        // Else, delete the node
 
        // node with only one child or no child
        if ((root->left == NULL) || (root->right == NULL)) {
            struct node* temp = root->left ? root->left : root->right;
 
            // No child case
            if (temp == NULL) {
                temp = root;
                root = NULL;
            }
            else // One child case
                *root = *temp; // Copy the contents of the non-empty child
 
            free(temp);
        }
        else {
            // node with two children: Get the inorder successor (smallest
            // in the right subtree)
            struct node* temp = minValueNode(root->right);
 
            // Copy the inorder successor's data to this node and update the count
            root->key = temp->key;
            root->count = temp->count;
            temp->count = 1;
 
            // Delete the inorder successor
            root->right = deleteNode(root->right, temp->key);
        }
    }
 
    // If the tree had only one node then return
    if (root == NULL)
        return root;
 
    // STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
    root->height = max(height(root->left), height(root->right)) + 1;
 
    // STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
    // this node became unbalanced)
    int balance = getBalance(root);
 
    // If this node becomes unbalanced, then there are 4 cases
 
    // Left Left Case
    if (balance > 1 && getBalance(root->left) >= 0)
        return rightRotate(root);
 
    // Left Right Case
    if (balance > 1 && getBalance(root->left) < 0) {
        root->left = leftRotate(root->left);
        return rightRotate(root);
    }
 
    // Right Right Case
    if (balance < -1 && getBalance(root->right) <= 0)
        return leftRotate(root);
 
    // Right Left Case
    if (balance < -1 && getBalance(root->right) > 0) {
        root->right = rightRotate(root->right);
        return leftRotate(root);
    }
 
    return root;
}
 
// A utility function to print preorder traversal of the tree.
// The function also prints height of every node
void preOrder(struct node* root)
{
    if (root != NULL) {
        cout << root->key << "("<<root->count << ")"<< " ";
        preOrder(root->left);
        preOrder(root->right);
    }
}
 
/* Driver program to test above function*/
int main()
{
    struct node* root = NULL;
 
    /* Constructing tree given in the above figure */
    root = insert(root, 9);
    root = insert(root, 5);
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 9);
    root = insert(root, 7);
    root = insert(root, 17);
 
    cout <<"Pre order traversal of the constructed AVL tree is \n";
    preOrder(root);
 
     
 
    cout <<"\nPre order traversal after deletion of 9 \n";
    preOrder(root);
     
    return 0;
}
 
// this code is contributed by shivanisinghss2110

C

// C++ program of AVL tree that
// handles duplicates
#include <stdio.h>
#include <stdlib.h>
 
// An AVL tree node
struct node {
    int key;
    struct node* left;
    struct node* right;
    int height;
    int count;
};
 
// A utility function to get maximum of two integers
int max(int a, int b);
 
// A utility function to get height of the tree
int height(struct node* N)
{
    if (N == NULL)
        return 0;
    return N->height;
}
 
// A utility function to get maximum of two integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
/* Helper function that allocates a new node with the given key and
    NULL left and right pointers. */
struct node* newNode(int key)
{
    struct node* node = (struct node*)
        malloc(sizeof(struct node));
    node->key = key;
    node->left = NULL;
    node->right = NULL;
    node->height = 1; // new node is initially added at leaf
    node->count = 1;
    return (node);
}
 
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct node* rightRotate(struct node* y)
{
    struct node* x = y->left;
    struct node* T2 = x->right;
 
    // Perform rotation
    x->right = y;
    y->left = T2;
 
    // Update heights
    y->height = max(height(y->left), height(y->right)) + 1;
    x->height = max(height(x->left), height(x->right)) + 1;
 
    // Return new root
    return x;
}
 
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct node* leftRotate(struct node* x)
{
    struct node* y = x->right;
    struct node* T2 = y->left;
 
    // Perform rotation
    y->left = x;
    x->right = T2;
 
    // Update heights
    x->height = max(height(x->left), height(x->right)) + 1;
    y->height = max(height(y->left), height(y->right)) + 1;
 
    // Return new root
    return y;
}
 
// Get Balance factor of node N
int getBalance(struct node* N)
{
    if (N == NULL)
        return 0;
    return height(N->left) - height(N->right);
}
 
struct node* insert(struct node* node, int key)
{
    /* 1.  Perform the normal BST rotation */
    if (node == NULL)
        return (newNode(key));
 
    // If key already exists in BST, increment count and return
    if (key == node->key) {
        (node->count)++;
        return node;
    }
 
    /* Otherwise, recur down the tree */
    if (key < node->key)
        node->left = insert(node->left, key);
    else
        node->right = insert(node->right, key);
 
    /* 2. Update height of this ancestor node */
    node->height = max(height(node->left), height(node->right)) + 1;
 
    /* 3. Get the balance factor of this ancestor node to check whether
       this node became unbalanced */
    int balance = getBalance(node);
 
    // If this node becomes unbalanced, then there are 4 cases
 
    // Left Left Case
    if (balance > 1 && key < node->left->key)
        return rightRotate(node);
 
    // Right Right Case
    if (balance < -1 && key > node->right->key)
        return leftRotate(node);
 
    // Left Right Case
    if (balance > 1 && key > node->left->key) {
        node->left = leftRotate(node->left);
        return rightRotate(node);
    }
 
    // Right Left Case
    if (balance < -1 && key < node->right->key) {
        node->right = rightRotate(node->right);
        return leftRotate(node);
    }
 
    /* return the (unchanged) node pointer */
    return node;
}
 
/* Given a non-empty binary search tree, return the node with minimum
   key value found in that tree. Note that the entire tree does not
   need to be searched. */
struct node* minValueNode(struct node* node)
{
    struct node* current = node;
 
    /* loop down to find the leftmost leaf */
    while (current->left != NULL)
        current = current->left;
 
    return current;
}
 
struct node* deleteNode(struct node* root, int key)
{
    // STEP 1: PERFORM STANDARD BST DELETE
 
    if (root == NULL)
        return root;
 
    // If the key to be deleted is smaller than the root's key,
    // then it lies in left subtree
    if (key < root->key)
        root->left = deleteNode(root->left, key);
 
    // If the key to be deleted is greater than the root's key,
    // then it lies in right subtree
    else if (key > root->key)
        root->right = deleteNode(root->right, key);
 
    // if key is same as root's key, then This is the node
    // to be deleted
    else {
        // If key is present more than once, simply decrement
        // count and return
        if (root->count > 1) {
            (root->count)--;
            return NULL;
        }
        // Else, delete the node
 
        // node with only one child or no child
        if ((root->left == NULL) || (root->right == NULL)) {
            struct node* temp = root->left ? root->left : root->right;
 
            // No child case
            if (temp == NULL) {
                temp = root;
                root = NULL;
            }
            else // One child case
                *root = *temp; // Copy the contents of the non-empty child
 
            free(temp);
        }
        else {
            // node with two children: Get the inorder successor (smallest
            // in the right subtree)
            struct node* temp = minValueNode(root->right);
 
            // Copy the inorder successor's data to this node and update the count
            root->key = temp->key;
            root->count = temp->count;
            temp->count = 1;
 
            // Delete the inorder successor
            root->right = deleteNode(root->right, temp->key);
        }
    }
 
    // If the tree had only one node then return
    if (root == NULL)
        return root;
 
    // STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
    root->height = max(height(root->left), height(root->right)) + 1;
 
    // STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
    // this node became unbalanced)
    int balance = getBalance(root);
 
    // If this node becomes unbalanced, then there are 4 cases
 
    // Left Left Case
    if (balance > 1 && getBalance(root->left) >= 0)
        return rightRotate(root);
 
    // Left Right Case
    if (balance > 1 && getBalance(root->left) < 0) {
        root->left = leftRotate(root->left);
        return rightRotate(root);
    }
 
    // Right Right Case
    if (balance < -1 && getBalance(root->right) <= 0)
        return leftRotate(root);
 
    // Right Left Case
    if (balance < -1 && getBalance(root->right) > 0) {
        root->right = rightRotate(root->right);
        return leftRotate(root);
    }
 
    return root;
}
 
// A utility function to print preorder traversal of the tree.
// The function also prints height of every node
void preOrder(struct node* root)
{
    if (root != NULL) {
        printf("%d(%d) ", root->key, root->count);
        preOrder(root->left);
        preOrder(root->right);
    }
}
 
/* Driver program to test above function*/
int main()
{
    struct node* root = NULL;
 
    /* Constructing tree given in the above figure */
    root = insert(root, 9);
    root = insert(root, 5);
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 9);
    root = insert(root, 7);
    root = insert(root, 17);
 
    printf("Pre order traversal of the constructed AVL tree is \n");
    preOrder(root);
 
    root = deleteNode(root, 9);
 
    printf("\nPre order traversal after deletion of 9 \n");
    preOrder(root);
 
    return 0;
}

Java

// Java program of AVL tree that handles duplicates
import java.util.*;
 
class solution {
 
    // An AVL tree node
    static class node {
        int key;
        node left;
        node right;
        int height;
        int count;
    }
 
    // A utility function to get height of the tree
    static int height(node N)
    {
        if (N == null)
            return 0;
        return N.height;
    }
 
    // A utility function to get maximum of two integers
    static int max(int a, int b)
    {
        return (a > b) ? a : b;
    }
 
    /* Helper function that allocates a new node with the given key and 
    null left and right pointers. */
    static node newNode(int key)
    {
        node node = new node();
        node.key = key;
        node.left = null;
        node.right = null;
        node.height = 1; // new node is initially added at leaf
        node.count = 1;
        return (node);
    }
 
    // A utility function to right rotate subtree rooted with y
    // See the diagram given above.
    static node rightRotate(node y)
    {
        node x = y.left;
        node T2 = x.right;
 
        // Perform rotation
        x.right = y;
        y.left = T2;
 
        // Update heights
        y.height = max(height(y.left), height(y.right)) + 1;
        x.height = max(height(x.left), height(x.right)) + 1;
 
        // Return new root
        return x;
    }
 
    // A utility function to left rotate subtree rooted with x
    // See the diagram given above.
    static node leftRotate(node x)
    {
        node y = x.right;
        node T2 = y.left;
 
        // Perform rotation
        y.left = x;
        x.right = T2;
 
        // Update heights
        x.height = max(height(x.left), height(x.right)) + 1;
        y.height = max(height(y.left), height(y.right)) + 1;
 
        // Return new root
        return y;
    }
 
    // Get Balance factor of node N
    static int getBalance(node N)
    {
        if (N == null)
            return 0;
        return height(N.left) - height(N.right);
    }
 
    static node insert(node node, int key)
    {
        /*1.  Perform the normal BST rotation */
        if (node == null)
            return (newNode(key));
 
        // If key already exists in BST, increment count and return
        if (key == node.key) {
            (node.count)++;
            return node;
        }
 
        /* Otherwise, recur down the tree */
        if (key < node.key)
            node.left = insert(node.left, key);
        else
            node.right = insert(node.right, key);
 
        /* 2. Update height of this ancestor node */
        node.height = max(height(node.left), height(node.right)) + 1;
 
        /* 3. Get the balance factor of this ancestor node to check whether 
       this node became unbalanced */
        int balance = getBalance(node);
 
        // If this node becomes unbalanced, then there are 4 cases
 
        // Left Left Case
        if (balance > 1 && key < node.left.key)
            return rightRotate(node);
 
        // Right Right Case
        if (balance < -1 && key > node.right.key)
            return leftRotate(node);
 
        // Left Right Case
        if (balance > 1 && key > node.left.key) {
            node.left = leftRotate(node.left);
            return rightRotate(node);
        }
 
        // Right Left Case
        if (balance < -1 && key < node.right.key) {
            node.right = rightRotate(node.right);
            return leftRotate(node);
        }
 
        /* return the (unchanged) node pointer */
        return node;
    }
 
    /* Given a non-empty binary search tree, return the node with minimum 
   key value found in that tree. Note that the entire tree does not 
   need to be searched. */
    static node minValueNode(node node)
    {
        node current = node;
 
        /* loop down to find the leftmost leaf */
        while (current.left != null)
            current = current.left;
 
        return current;
    }
 
    static node deleteNode(node root, int key)
    {
        // STEP 1: PERFORM STANDARD BST DELETE
 
        if (root == null)
            return root;
 
        // If the key to be deleted is smaller than the root's key,
        // then it lies in left subtree
        if (key < root.key)
            root.left = deleteNode(root.left, key);
 
        // If the key to be deleted is greater than the root's key,
        // then it lies in right subtree
        else if (key > root.key)
            root.right = deleteNode(root.right, key);
 
        // if key is same as root's key, then This is the node
        // to be deleted
        else {
            // If key is present more than once, simply decrement
            // count and return
            if (root.count > 1) {
                (root.count)--;
                return null;
            }
            // ElSE, delete the node
 
            // node with only one child or no child
            if ((root.left == null) || (root.right == null)) {
                node temp = root.left != null ? root.left : root.right;
 
                // No child case
                if (temp == null) {
                    temp = root;
                    root = null;
                }
                else // One child case
                    root = temp; // Copy the contents of the non-empty child
            }
            else {
                // node with two children: Get the inorder successor (smallest
                // in the right subtree)
                node temp = minValueNode(root.right);
 
                // Copy the inorder successor's data to this node and update the count
                root.key = temp.key;
                root.count = temp.count;
                temp.count = 1;
 
                // Delete the inorder successor
                root.right = deleteNode(root.right, temp.key);
            }
        }
 
        // If the tree had only one node then return
        if (root == null)
            return root;
 
        // STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
        root.height = max(height(root.left), height(root.right)) + 1;
 
        // STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
        // this node became unbalanced)
        int balance = getBalance(root);
 
        // If this node becomes unbalanced, then there are 4 cases
 
        // Left Left Case
        if (balance > 1 && getBalance(root.left) >= 0)
            return rightRotate(root);
 
        // Left Right Case
        if (balance > 1 && getBalance(root.left) < 0) {
            root.left = leftRotate(root.left);
            return rightRotate(root);
        }
 
        // Right Right Case
        if (balance < -1 && getBalance(root.right) <= 0)
            return leftRotate(root);
 
        // Right Left Case
        if (balance < -1 && getBalance(root.right) > 0) {
            root.right = rightRotate(root.right);
            return leftRotate(root);
        }
 
        return root;
    }
 
    // A utility function to print preorder traversal of the tree.
    // The function also prints height of every node
    static void preOrder(node root)
    {
        if (root != null) {
            System.out.printf("%d(%d) ", root.key, root.count);
            preOrder(root.left);
            preOrder(root.right);
        }
    }
 
    /* Driver program to test above function*/
    public static void main(String args[])
    {
        node root = null;
 
        /* Constructing tree given in the above figure */
        root = insert(root, 9);
        root = insert(root, 5);
        root = insert(root, 10);
        root = insert(root, 5);
        root = insert(root, 9);
        root = insert(root, 7);
        root = insert(root, 17);
 
        System.out.printf("Pre order traversal of the constructed AVL tree is \n");
        preOrder(root);
 
        deleteNode(root, 9);
 
        System.out.printf("\nPre order traversal after deletion of 9 \n");
        preOrder(root);
    }
}
// contributed by Arnab Kundu

C#

// C# program of AVL tree that
// handles duplicates
using System;
 
class GFG {
 
    // An AVL tree node
    class node {
        public int key;
        public node left;
        public node right;
        public int height;
        public int count;
    }
 
    // A utility function to get
    // height of the tree
    static int height(node N)
    {
        if (N == null)
            return 0;
        return N.height;
    }
 
    // A utility function to get
    // maximum of two integers
    static int max(int a, int b)
    {
        return (a > b) ? a : b;
    }
 
    /* Helper function that allocates a 
    new node with the given key and 
    null left and right pointers. */
    static node newNode(int key)
    {
        node node = new node();
        node.key = key;
        node.left = null;
        node.right = null;
        node.height = 1; // new node is initially
        // added at leaf
        node.count = 1;
        return (node);
    }
 
    // A utility function to right
    // rotate subtree rooted with y
    // See the diagram given above.
    static node rightRotate(node y)
    {
        node x = y.left;
        node T2 = x.right;
 
        // Perform rotation
        x.right = y;
        y.left = T2;
 
        // Update heights
        y.height = max(height(y.left),
                       height(y.right))
                   + 1;
        x.height = max(height(x.left),
                       height(x.right))
                   + 1;
 
        // Return new root
        return x;
    }
 
    // A utility function to left
    // rotate subtree rooted with x
    // See the diagram given above.
    static node leftRotate(node x)
    {
        node y = x.right;
        node T2 = y.left;
 
        // Perform rotation
        y.left = x;
        x.right = T2;
 
        // Update heights
        x.height = max(height(x.left),
                       height(x.right))
                   + 1;
        y.height = max(height(y.left),
                       height(y.right))
                   + 1;
 
        // Return new root
        return y;
    }
 
    // Get Balance factor of node N
    static int getBalance(node N)
    {
        if (N == null)
            return 0;
        return height(N.left) - height(N.right);
    }
 
    static node insert(node node, int key)
    {
        /*1. Perform the normal BST rotation */
        if (node == null)
            return (newNode(key));
 
        // If key already exists in BST,
        // increment count and return
        if (key == node.key) {
            (node.count)++;
            return node;
        }
 
        /* Otherwise, recur down the tree */
        if (key < node.key)
            node.left = insert(node.left, key);
        else
            node.right = insert(node.right, key);
 
        /* 2. Update height of this 
          ancestor node */
        node.height = max(height(node.left),
                          height(node.right))
                      + 1;
 
        /* 3. Get the balance factor of 
    this ancestor node to check whether 
    this node became unbalanced */
        int balance = getBalance(node);
 
        // If this node becomes unbalanced,
        // then there are 4 cases
 
        // Left Left Case
        if (balance > 1 && key < node.left.key)
            return rightRotate(node);
 
        // Right Right Case
        if (balance < -1 && key > node.right.key)
            return leftRotate(node);
 
        // Left Right Case
        if (balance > 1 && key > node.left.key) {
            node.left = leftRotate(node.left);
            return rightRotate(node);
        }
 
        // Right Left Case
        if (balance < -1 && key < node.right.key) {
            node.right = rightRotate(node.right);
            return leftRotate(node);
        }
 
        /* return the (unchanged)
       node pointer */
        return node;
    }
 
    /* Given a non-empty binary search 
   tree, return the node with minimum 
   key value found in that tree. Note 
   that the entire tree does not 
   need to be searched. */
    static node minValueNode(node node)
    {
        node current = node;
 
        /* loop down to find the
       leftmost leaf */
        while (current.left != null)
            current = current.left;
 
        return current;
    }
 
    static node deleteNode(node root, int key)
    {
        // STEP 1: PERFORM STANDARD BST DELETE
        if (root == null)
            return root;
 
        // If the key to be deleted is
        // smaller than the root's key,
        // then it lies in left subtree
        if (key < root.key)
            root.left = deleteNode(root.left, key);
 
        // If the key to be deleted is
        // greater than the root's key,
        // then it lies in right subtree
        else if (key > root.key)
            root.right = deleteNode(root.right, key);
 
        // if key is same as root's key,
        // then this is the node to be deleted
        else {
            // If key is present more than
            // once, simply decrement
            // count and return
            if (root.count > 1) {
                (root.count)--;
                return null;
            }
 
            // ElSE, delete the node
 
            // node with only one child
            // or no child
            if ((root.left == null) || (root.right == null)) {
                node temp = root.left != null ? root.left : root.right;
 
                // No child case
                if (temp == null) {
                    temp = root;
                    root = null;
                }
                else // One child case
                    root = temp; // Copy the contents of
                // the non-empty child
            }
            else {
                // node with two children: Get
                // the inorder successor (smallest
                // in the right subtree)
                node temp = minValueNode(root.right);
 
                // Copy the inorder successor's
                // data to this node and update the count
                root.key = temp.key;
                root.count = temp.count;
                temp.count = 1;
 
                // Delete the inorder successor
                root.right = deleteNode(root.right,
                                        temp.key);
            }
        }
 
        // If the tree had only one
        // node then return
        if (root == null)
            return root;
 
        // STEP 2: UPDATE HEIGHT OF
        // THE CURRENT NODE
        root.height = max(height(root.left),
                          height(root.right))
                      + 1;
 
        // STEP 3: GET THE BALANCE FACTOR
        // OF THIS NODE (to check whether
        // this node became unbalanced)
        int balance = getBalance(root);
 
        // If this node becomes unbalanced,
        // then there are 4 cases
 
        // Left Left Case
        if (balance > 1 && getBalance(root.left) >= 0)
            return rightRotate(root);
 
        // Left Right Case
        if (balance > 1 && getBalance(root.left) < 0) {
            root.left = leftRotate(root.left);
            return rightRotate(root);
        }
 
        // Right Right Case
        if (balance < -1 && getBalance(root.right) <= 0)
            return leftRotate(root);
 
        // Right Left Case
        if (balance < -1 && getBalance(root.right) > 0) {
            root.right = rightRotate(root.right);
            return leftRotate(root);
        }
 
        return root;
    }
 
    // A utility function to print
    // preorder traversal of the tree.
    // The function also prints height
    // of every node
    static void preOrder(node root)
    {
        if (root != null) {
            Console.Write(root.key + "(" + root.count + ") ");
            preOrder(root.left);
            preOrder(root.right);
        }
    }
 
    // Driver Code
    static public void Main(String[] args)
    {
        node root = null;
 
        /* Constructing tree given in 
       the above figure */
        root = insert(root, 9);
        root = insert(root, 5);
        root = insert(root, 10);
        root = insert(root, 5);
        root = insert(root, 9);
        root = insert(root, 7);
        root = insert(root, 17);
 
        Console.Write("Pre order traversal of "
                      + "the constructed AVL tree is \n");
        preOrder(root);
 
        deleteNode(root, 9);
 
        Console.Write("\nPre order traversal after "
                      + "deletion of 9 \n");
        preOrder(root);
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python code to delete a node in AVL tree
# Generic tree node class
 
 
class TreeNode():
    def __init__(self, val):
        self.count = 1    # assigning count variable so that during insertion in will be incremented for duplicate values
        # and during deletion, it will be decremented if has multiple copies.
        self.height = 1
        self.val = val
        self.left = None
        self.right = None
# only insertion and deletion will be affected. if multiple copies are there, entry(count) will be printed during traversal.
 
# AVL tree class which supports insertion,
# deletion operations
 
 
class AVL_Tree(object):
 
    def insert(self, root, key):
 
        # Step 1 - Perform normal BST
        if not root:
            return TreeNode(key)
        else if key < root.val:
            root.left = self.insert(root.left, key)
        else if key > root.val:
            root.right = self.insert(root.right, key)
        else:
            root.count += 1  # incrementing count if same entry is inserted.
 
        # Step 2 - Update the height of the
        # ancestor node
        root.height = 1 + max(self.getHeight(root.left),
                              self.getHeight(root.right))
 
        # Step 3 - Get the balance factor
        balance = self.getBalance(root)
 
        # Step 4 - If the node is unbalanced,
        # then try out the 4 cases
        # Case 1 - Left Left
        if balance > 1 and key < root.left.val:
            return self.rightRotate(root)
 
        # Case 2 - Right Right
        if balance < -1 and key > root.right.val:
            return self.leftRotate(root)
 
        # Case 3 - Left Right
        if balance > 1 and key > root.left.val:
            root.left = self.leftRotate(root.left)
            return self.rightRotate(root)
 
        # Case 4 - Right Left
        if balance < -1 and key < root.right.val:
            root.right = self.rightRotate(root.right)
            return self.leftRotate(root)
 
        return root
 
    # Recursive function to delete a node with
    # given key from subtree with given root.
    # It returns root of the modified subtree.
    def delete(self, root, key):
 
        # Step 1 - Perform standard BST delete
        if not root:
            return root
 
        else if key < root.val:
            root.left = self.delete(root.left, key)
 
        else if key > root.val:
            root.right = self.delete(root.right, key)
 
        else:
            if root.count > 1:  # if count is more than one i.e multiple copies are there
                root.count -= 1  # just decrement count
                return root   # so that one copy will be deleted and return
 
            if root.left is None:
                temp = root.right
                root = None
                return temp
 
            else if root.right is None:
                temp = root.left
                root = None
                return temp
 
            temp = self.getMinValueNode(root.right)
            root.val = temp.val
            root.right = self.delete(root.right,
                                     temp.val)
 
        # If the tree has only one node,
        # simply return it
        if root is None:
            return root
 
        # Step 2 - Update the height of the
        # ancestor node
        root.height = 1 + max(self.getHeight(root.left),
                              self.getHeight(root.right))
 
        # Step 3 - Get the balance factor
        balance = self.getBalance(root)
 
        # Step 4 - If the node is unbalanced,
        # then try out the 4 cases
        # Case 1 - Left Left
        if balance > 1 and self.getBalance(root.left) >= 0:
            return self.rightRotate(root)
 
        # Case 2 - Right Right
        if balance < -1 and self.getBalance(root.right) <= 0:
            return self.leftRotate(root)
 
        # Case 3 - Left Right
        if balance > 1 and self.getBalance(root.left) < 0:
            root.left = self.leftRotate(root.left)
            return self.rightRotate(root)
 
        # Case 4 - Right Left
        if balance < -1 and self.getBalance(root.right) > 0:
            root.right = self.rightRotate(root.right)
            return self.leftRotate(root)
 
        return root
 
    def leftRotate(self, z):
 
        y = z.right
        T2 = y.left
 
        # Perform rotation
        y.left = z
        z.right = T2
 
        # Update heights
        z.height = 1 + max(self.getHeight(z.left),
                           self.getHeight(z.right))
        y.height = 1 + max(self.getHeight(y.left),
                           self.getHeight(y.right))
 
        # Return the new root
        return y
 
    def rightRotate(self, z):
 
        y = z.left
        T3 = y.right
 
        # Perform rotation
        y.right = z
        z.left = T3
 
        # Update heights
        z.height = 1 + max(self.getHeight(z.left),
                           self.getHeight(z.right))
        y.height = 1 + max(self.getHeight(y.left),
                           self.getHeight(y.right))
 
        # Return the new root
        return y
 
    def getHeight(self, root):
        if not root:
            return 0
 
        return root.height
 
    def getBalance(self, root):
        if not root:
            return 0
 
        return self.getHeight(root.left) - self.getHeight(root.right)
 
    def getMinValueNode(self, root):
        if root is None or root.left is None:
            return root
 
        return self.getMinValueNode(root.left)
 
    def preOrder(self, root):
 
        if not root:
            return
 
        print("{}({}) ".format(root.val, root.count), end="")
        self.preOrder(root.left)
        self.preOrder(root.right)
 
 
myTree = AVL_Tree()
root = None
nums = [9, 5, 10, 5, 9, 7, 17]
 
for num in nums:
    root = myTree.insert(root, num)
 
# Preorder Traversal
print("Preorder Traversal after insertion -")
myTree.preOrder(root)
print()
 
# Delete
key = 10
root = myTree.delete(root, key)
key = 10
root = myTree.delete(root, key)
key = -1
root = myTree.delete(root, key)
key = 0
root = myTree.delete(root, key)
 
# Preorder Traversal
print("Preorder Traversal after deletion -")
myTree.preOrder(root)
print()
 
# This code is contributed by Ajitesh Pathak

Javascript

<script>
 
    // JavaScript program of AVL tree that handles duplicates
     
    class Node
    {
        constructor() {
           this.left;
           this.right;
           this.key;
           this.height;
           this.count;
        }
    }
     
    // A utility function to get height of the tree
    function height(N)
    {
        if (N == null)
            return 0;
        return N.height;
    }
  
    // A utility function to get maximum of two integers
    function max(a, b)
    {
        return (a > b) ? a : b;
    }
  
    /* Helper function that allocates a new 
    node with the given key and
    null left and right pointers. */
    function newNode(key)
    {
        let node = new Node();
        node.key = key;
        node.left = null;
        node.right = null;
        node.height = 1; // new node is initially added at leaf
        node.count = 1;
        return (node);
    }
  
    // A utility function to right rotate
    // subtree rooted with y
    // See the diagram given above.
    function rightRotate(y)
    {
        let x = y.left;
        let T2 = x.right;
  
        // Perform rotation
        x.right = y;
        y.left = T2;
  
        // Update heights
        y.height = max(height(y.left), height(y.right)) + 1;
        x.height = max(height(x.left), height(x.right)) + 1;
  
        // Return new root
        return x;
    }
  
    // A utility function to left rotate subtree rooted with x
    // See the diagram given above.
    function leftRotate(x)
    {
        let y = x.right;
        let T2 = y.left;
  
        // Perform rotation
        y.left = x;
        x.right = T2;
  
        // Update heights
        x.height = max(height(x.left), height(x.right)) + 1;
        y.height = max(height(y.left), height(y.right)) + 1;
  
        // Return new root
        return y;
    }
  
    // Get Balance factor of node N
    function getBalance(N)
    {
        if (N == null)
            return 0;
        return height(N.left) - height(N.right);
    }
  
    function insert(node, key)
    {
        /*1.  Perform the normal BST rotation */
        if (node == null)
            return (newNode(key));
  
        // If key already exists in BST, increment count and return
        if (key == node.key) {
            (node.count)++;
            return node;
        }
  
        /* Otherwise, recur down the tree */
        if (key < node.key)
            node.left = insert(node.left, key);
        else
            node.right = insert(node.right, key);
  
        /* 2. Update height of this ancestor node */
        node.height = max(height(node.left), height(node.right)) + 1;
  
        /* 3. Get the balance factor of this 
        ancestor node to check whether
        this node became unbalanced */
        let balance = getBalance(node);
  
        // If this node becomes unbalanced, then there are 4 cases
  
        // Left Left Case
        if (balance > 1 && key < node.left.key)
            return rightRotate(node);
  
        // Right Right Case
        if (balance < -1 && key > node.right.key)
            return leftRotate(node);
  
        // Left Right Case
        if (balance > 1 && key > node.left.key) {
            node.left = leftRotate(node.left);
            return rightRotate(node);
        }
  
        // Right Left Case
        if (balance < -1 && key < node.right.key) {
            node.right = rightRotate(node.right);
            return leftRotate(node);
        }
  
        /* return the (unchanged) node pointer */
        return node;
    }
  
    /* Given a non-empty binary search tree,
       return the node with minimum
       key value found in that tree. Note that 
       the entire tree does not
       need to be searched. */
    function minValueNode(node)
    {
        let current = node;
  
        /* loop down to find the leftmost leaf */
        while (current.left != null)
            current = current.left;
  
        return current;
    }
  
    function deleteNode(root, key)
    {
        // STEP 1: PERFORM STANDARD BST DELETE
  
        if (root == null)
            return root;
  
        // If the key to be deleted is smaller than the root's key,
        // then it lies in left subtree
        if (key < root.key)
            root.left = deleteNode(root.left, key);
  
        // If the key to be deleted is greater than the root's key,
        // then it lies in right subtree
        else if (key > root.key)
            root.right = deleteNode(root.right, key);
  
        // if key is same as root's key, then This is the node
        // to be deleted
        else {
            // If key is present more than once, simply decrement
            // count and return
            if (root.count > 1) {
                (root.count)--;
                return null;
            }
            // ElSE, delete the node
  
            // node with only one child or no child
            if ((root.left == null) || (root.right == null)) {
                let temp = root.left != null ? root.left : root.right;
  
                // No child case
                if (temp == null) {
                    temp = root;
                    root = null;
                }
                else // One child case
                // Copy the contents of the non-empty child
                    root = temp; 
            }
            else {
                // node with two children: Get the
                // inorder successor (smallest
                // in the right subtree)
                let temp = minValueNode(root.right);
  
                // Copy the inorder successor's data to
                // this node and update the count
                root.key = temp.key;
                root.count = temp.count;
                temp.count = 1;
  
                // Delete the inorder successor
                root.right = deleteNode(root.right, temp.key);
            }
        }
  
        // If the tree had only one node then return
        if (root == null)
            return root;
  
        // STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
        root.height = max(height(root.left), height(root.right)) + 1;
  
        // STEP 3: GET THE BALANCE FACTOR 
        // OF THIS NODE (to check whether
        // this node became unbalanced)
        let balance = getBalance(root);
  
        // If this node becomes unbalanced, then there are 4 cases
  
        // Left Left Case
        if (balance > 1 && getBalance(root.left) >= 0)
            return rightRotate(root);
  
        // Left Right Case
        if (balance > 1 && getBalance(root.left) < 0) {
            root.left = leftRotate(root.left);
            return rightRotate(root);
        }
  
        // Right Right Case
        if (balance < -1 && getBalance(root.right) <= 0)
            return leftRotate(root);
  
        // Right Left Case
        if (balance < -1 && getBalance(root.right) > 0) {
            root.right = rightRotate(root.right);
            return leftRotate(root);
        }
  
        return root;
    }
  
    // A utility function to print preorder traversal of the tree.
    // The function also prints height of every node
    function preOrder(root)
    {
        if (root != null) {
            document.write(root.key + "(" + root.count + ") ");
            preOrder(root.left);
            preOrder(root.right);
        }
    }
     
    let root = null;
  
    /* Constructing tree given in the above figure */
    root = insert(root, 9);
    root = insert(root, 5);
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 9);
    root = insert(root, 7);
    root = insert(root, 17);
 
    document.write(
    "Pre order traversal of the constructed AVL tree is " + 
    "</br>");
    preOrder(root);
 
    deleteNode(root, 9);
 
    document.write(
    "</br>" + "Pre order traversal after deletion of 9 " + 
    "</br>");
    preOrder(root);
     
</script>

Output

Pre order traversal of the constructed AVL tree is 
9(2) 5(2) 7(1) 10(1) 17(1) 
Pre order traversal after deletion of 9 
9(2) 5(2) 7(1) 10(1) 17(1)

Time Complexity: O(N) ,here ‘N’ is for length of the Linked list.

Auxiliary Space: O(1), since no extra space required.

Thanks to Rounaq Jhunjhunu Wala for sharing initial code.

GeeksforGeeks

Improve

Practice questions on Height balanced/AVL Tree

Count greater nodes in AVL tree

AVL with duplicate keys

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Java

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Python3

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